HDU6027(快速幂)（水）

Easy Summation

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1329    Accepted Submission(s): 538

Problem Description

You are encountered with a traditional problem concerning the sums of powers.
Given two integers n and k. Let f(i)=ik, please evaluate the sum f(1)+f(2)+...+f(n). The problem is simple as it looks, apart from the value of n in this question is quite large.
Can you figure the answer out? Since the answer may be too large, please output the answer modulo 109+7.

 

Input

The first line of the input contains an integer T(1≤T≤20), denoting the number of test cases.
Each of the following T lines contains two integers n(1≤n≤10000) and k(0≤k≤5).

 

Output

For each test case, print a single line containing an integer modulo 109+7.

 

Sample Input

32 54 24 1

 

Sample Output

333010

 

#include<cstdio>#include<iostream>#include<cstring>#include<string.h>using namespace std;typedef long long ll;int n,k;ll ans;const ll MOD=1e9+7;ll fsm(ll a,int b){ll res=1;while(b>0){if(b&1)res=(res*a)%MOD;a=(a*a)%MOD;b>>=1;}return res;}int main(){int t;scanf("%d",&t);while(t--){ans=0;scanf("%d%d",&n,&k);for(int i=1;i<=n;i++){ans=(ans+fsm(i,k))%MOD;}printf("%lld\n",ans);}}